scipy.linalg.solve_discrete_lyapunov¶
-
scipy.linalg.
solve_discrete_lyapunov
(a, q, method=None)[source]¶ Solves the discrete Lyapunov equation \(AXA^H - X + Q = 0\).
Parameters: a, q : (M, M) array_like
Square matrices corresponding to A and Q in the equation above respectively. Must have the same shape.
method : {‘direct’, ‘bilinear’}, optional
Type of solver.
If not given, chosen to be
direct
ifM
is less than 10 andbilinear
otherwise.Returns: x : ndarray
Solution to the discrete Lyapunov equation
See also
solve_continuous_lyapunov
- computes the solution to the continuous-time Lyapunov equation
Notes
This section describes the available solvers that can be selected by the ‘method’ parameter. The default method is direct if
M
is less than 10 andbilinear
otherwise.Method direct uses a direct analytical solution to the discrete Lyapunov equation. The algorithm is given in, for example, [R155]. However it requires the linear solution of a system with dimension \(M^2\) so that performance degrades rapidly for even moderately sized matrices.
Method bilinear uses a bilinear transformation to convert the discrete Lyapunov equation to a continuous Lyapunov equation \((BX+XB'=-C)\) where \(B=(A-I)(A+I)^{-1}\) and \(C=2(A' + I)^{-1} Q (A + I)^{-1}\). The continuous equation can be efficiently solved since it is a special case of a Sylvester equation. The transformation algorithm is from Popov (1964) as described in [R156].
New in version 0.11.0.
References
[R155] (1, 2) Hamilton, James D. Time Series Analysis, Princeton: Princeton University Press, 1994. 265. Print. http://doc1.lbfl.li/aca/FLMF037168.pdf [R156] (1, 2) Gajic, Z., and M.T.J. Qureshi. 2008. Lyapunov Matrix Equation in System Stability and Control. Dover Books on Engineering Series. Dover Publications. Examples
Given a and q solve for x:
>>> from scipy import linalg >>> a = np.array([[0.2, 0.5],[0.7, -0.9]]) >>> q = np.eye(2) >>> x = linalg.solve_discrete_lyapunov(a, q) >>> x array([[ 0.70872893, 1.43518822], [ 1.43518822, -2.4266315 ]]) >>> np.allclose(a.dot(x).dot(a.T)-x, -q) True